Answer :
The speed of the motorbike must be 13.3 m/s
Explanation:
There are two forces acting on the motorcycle during its circular motion:
- The force of gravity, of magnitude [tex]mg[/tex] (where m is the mass of the motorbike+rider and [tex]g=9.8 m/s^2[/tex] is the acceleration of gravity), always pointing downward
- The normal reaction of the track on the motorbike, N
When the rider is at the top of the loop, the normal reaction points downward. The sum of the two forces must be equal to the net force, which is the centripetal force, therefore we can write:
[tex]mg+N = m\frac{v^2}{r}[/tex]
where
v is the speed of the motorbike
r = 18 m is the radius of the loop
Here we want to find the minimum speed that the motorbike must have in order not to lose contact with the track: this means, the speed at which the normal reaction is zero,
N = 0
Substituting and solving for v, we find:
[tex]v=\sqrt{gr}=\sqrt{(9.8)(18)}=13.3 m/s[/tex]
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In this case, the speed of the motorbike must be - 13.3 m/s
Centripetal force
In order for the rider to pass the top of the loop without falling, his weight must be equal to the centripetal force:
There are two forces works on the motorcycle during its circular motion:
- The force of gravity, of magnitude:
= mg
where, m = mass of the motorbike and rider
g = acceleration of gravity always pointing downward
- The normal reaction of the track on the motorbike, N
The total force or net force must be equal to both given forces, which is the centripetal force, therefore we can write:
[tex]\frac{mv^{2}}{r} = mg[/tex]
[tex]v =\sqrt{gr}[/tex]
where
v = the speed of the motorbike
r = 18 m is the radius of the loop
[tex]v = \sqrt{(9.8) (18)}[/tex]
[tex]v= 13.3[/tex]
Learn more about circular motion:
brainly.com/question/2562955